---
title: "ACSR 132 kV line: parameters and charging MVA"
subject: "Power System Analysis I"
chapter: 4
key: "semester_4/power_system_analysis_i:acsr-132kv-8m-parameters-charging"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/acsr-132kv-8m-parameters-charging
site: BCT Engineering
---

# ACSR 132 kV line: parameters and charging MVA

Power System Analysis I · Chapter 4: Line Parameter Calculations

- Asked in: 2081 Chaitra (new course), 3+3 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/acsr-132kv-8m-parameters-charging
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-4

## Question

A 3-phase 132 kV, 50 Hz line has ACSR conductors of equivalent copper area 1.5 cm² and effective diameter 39.8 mm, spaced equilaterally 8 m apart. (i) Find the line parameters (ii) find the charging current and charging MVA. Resistivity of copper is 1.73×10⁻⁶ Ω-cm.

## Short answer

Resistance R = ρl/A = 0.1153 Ω/km. Inductance L = 2×10⁻⁷ ln(D/r') = 1.249 mH/km (XL = 0.3925 Ω/km). Capacitance C = 2πε₀/ln(D/r) = 0.00928 μF/km. Charging current Ic = Vph/Xc = 0.222 A/km, and three-phase charging MVA = √3 VLL Ic = 0.0508 MVAR/km.

## Answer

The line length is not given, so all results are **per km**. The ACSR conductor is treated as a solid round conductor of the given effective diameter, with GMR $r' = 0.7788r$ (no strand data is given).

#### Data
- $V_{LL} = 132$ kV, so $V_{ph} = 132/\sqrt3 = 76.21$ kV; f = 50 Hz, $\omega = 314.16$ rad/s
- $r = 39.8/2 = 19.9$ mm $= 1.99$ cm; D = 8 m = 800 cm (equilateral)
- Copper-equivalent area $A = 1.5$ cm²; $\rho = 1.73\times10^{-6}\ \Omega$-cm

#### (i) Line parameters
**Resistance** (l = 1 km = $10^5$ cm):
$$
R = \frac{\rho l}{A} = \frac{1.73\times10^{-6}\times10^5}{1.5} = 0.1153\ \Omega/\text{km}
$$

**Inductance:** $r' = 0.7788\times1.99 = 1.5498$ cm.
$$
\begin{aligned}
L &= 2\times10^{-7}\ln\frac{D}{r'} = 2\times10^{-7}\ln\frac{800}{1.5498}\\
&= 2\times10^{-7}\times6.2465 = 1.2493\ \text{mH/km}\\
X_L &= 314.16\times1.2493\times10^{-3} = 0.3925\ \Omega/\text{km}
\end{aligned}
$$

**Capacitance** to neutral:
$$
\begin{aligned}
C &= \frac{2\pi\varepsilon_0}{\ln(D/r)} = \frac{5.563\times10^{-11}}{\ln(800/1.99)} = \frac{5.563\times10^{-11}}{5.9965}\\
&= 9.277\times10^{-12}\ \text{F/m} = 9.277\times10^{-3}\ \mu\text{F/km}\\
X_C &= \frac{1}{\omega C} = \frac{1}{314.16\times9.277\times10^{-9}} = 3.431\times10^5\ \Omega\cdot\text{km}
\end{aligned}
$$

#### (ii) Charging current and charging MVA
$$
\begin{aligned}
I_c &= \frac{V_{ph}}{X_C} = 76210\times314.16\times9.277\times10^{-9} = 0.2221\ \text{A/km}\\
Q_c &= 3V_{ph}I_c = \sqrt3\times132\times10^3\times0.2221\\
&= 50.78\ \text{kVAR/km} = 0.0508\ \text{MVAR/km}
\end{aligned}
$$

| Parameter | Value per phase per km |
|---|---|
| R | 0.1153 Ω |
| L ($X_L$) | 1.249 mH (0.3925 Ω) |
| C ($X_C$) | 0.009277 μF (343.1 kΩ·km) |
| Charging current | 0.2221 A |
| Charging MVA (3-phase) | 0.0508 MVAR |

**Answer:** R = 0.115 Ω/km, L = 1.249 mH/km, C = 0.00928 μF/km; charging current 0.222 A/km and charging MVA 0.0508 MVAR/km. For example, a 100 km line would draw 22.2 A and 5.08 MVAR.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is the resistance per km found from the given copper-equivalent area?
   A. R = rho l / A = 1.73e-6 x 1e5 / 1.5 = 0.1153 ohm/km
   B. R = rho A / l
   C. R = rho l x A = 0.26 ohm/km
   D. R = 1/(rho A)
   Answer: A. R = rho l / A with l = 1 km = 1e5 cm gives 0.1153 ohm/km. (Key point: Resistance)

2. Which GMR is used in the inductance formula for the solid-treated conductor?
   A. r' = 0.7788 r
   B. r' = r
   C. r' = 1.09 r
   D. r' = 2 r
   Answer: A. Inductance uses GMR 0.7788 r; here r' = 1.55 cm and L = 1.249 mH/km. (Key point: Inductance formula)

3. Which radius is used in the capacitance formula C = 2 pi eps0 / ln(D/r)?
   A. The actual radius r = 1.99 cm
   B. The GMR 0.7788 r
   C. The diameter 3.98 cm
   D. The spacing D
   Answer: A. Capacitance depends on the physical conductor radius, not the GMR. (Key point: Capacitance radius)

4. How is the charging current per km obtained, and what is its value?
   A. Ic = V_ph w C = 0.222 A/km
   B. Ic = V_LL / X_L = 336 A/km
   C. Ic = V_ph / R = 661 kA/km
   D. Ic = V_LL w C = 0.385 A/km
   Answer: A. Ic = V_ph x w x C = 76210 x 314.16 x 9.277e-9 = 0.222 A/km. (Key point: Charging current)

## Related in Power System Analysis I

- [Charging current of 60 Hz horizontal 132 kV line](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Ahorizontal-60hz-132kv-charging) — Power System Analysis I, chapter 4
- [Charging current of 132 kV line, 6 m equilateral](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Aequilateral-132kv-6m-charging) — Power System Analysis I, chapter 4
- [132 kV triangular line: L, C, charging MVAR](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Atriangular-132kv-18mm-parameters) — Power System Analysis I, chapter 4
- [Charging VAR of 132 kV equilateral line](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Acharging-var-132kv-6m-1-1cm) — Power System Analysis I, chapter 4
- [Nominal-T 100 km line, 50 MVA at 132 kV](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Anominal-t-100km-50mva) — Power System Analysis I, chapter 6
- [132 kV horizontal line: susceptance and charging](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Ahorizontal-132kv-4-5m-susceptance) — Power System Analysis I, chapter 4

## Related in other subjects

- [50 kVA transformer: equivalent parameters, copper loss](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Atf-50kva-4400-220-equivalent-resistance) — Electrical Machines, chapter 2
- [Conductor selection, 100 MW 70 km 132 kV](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Aconductor-selection-100mw-70km-132kv) — Transmission and Distribution System Design, chapter 4
- [Equivalent impedance and copper loss of 50 kVA transformer](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3A50kva-4400-220-equivalent-copper-loss) — Electrical Machines I, chapter 2
- [Air clearances for 132 kV double circuit line](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Aair-clearance-132kv-double-circuit) — Transmission and Distribution System Design, chapter 4